MANE 4240 & CIVL 4240
Introduction to Finite Elements
Reading assignment:
Logan 10.2 + Lecture notes
Summary:
Prof. Suvranu De
Four-noded
rectangular element
Finite element formulation for 2D:
Step 1: Divide the body into finite elements connected to each
other through special points (“nodes”)
py
v3
3
px
4
3
u3
⎧u 1 ⎫
v4
2
⎪v ⎪
v2
Element ‘e’
v
⎪ 1⎪
1
4
⎪u 2 ⎪
u
u4
ST
u2
⎪ ⎪
v1
2
⎪v 2 ⎪
y
d=⎨ ⎬
x
y
⎪u 3 ⎪
Su
u1
⎪v 3 ⎪
1
⎪ ⎪
v
x
x
⎪u 4 ⎪
u
⎪v ⎪
⎩ 4⎭
• Computation of shape functions for 4-noded quad
• Special case: rectangular element
• Properties of shape functions
• Computation of strain-displacement matrix
• Example problem
•Hint at how to generate shape functions of higher order
(Lagrange) elements
Summary: For each element
Displacement approximation in terms of shape functions
u=Nd
Strain approximation in terms of strain-displacement matrix
ε=Bd
Stress approximation
σ = DB d
Element stiffness matrix
k = ∫ e B D B dV
T
V
Element nodal load vector
f = ∫ e N X dV + ∫ e N T S dS
V
ST
T
f
T
b
f
S
1
Constant Strain Triangle (CST) : Simplest 2D finite element
Formula for the shape functions are
v1
v3
1
u1
(x1,y1)
v1
1
(x1,y1)
v2
y
u1
v
(x,y)
u
v3
y
(x3,y3)
3
u2
2 (x2,y2)
v2
x
⎧ u1 ⎫
⎪v ⎪
⎪ 1⎪
0 ⎤ ⎪⎪u 2 ⎪⎪
⎨ ⎬
N 3 ⎥⎦ ⎪v 2 ⎪
⎪u 3 ⎪
⎪ ⎪
⎩⎪ v 3 ⎭⎪
u=Nd
0
N2
0
N3
N1
0
N2
0
Approximation of the strains
⎧ ε x
⎪
ε = ⎨ε y
⎪γ
⎩ xy
∂u
⎧
⎫
⎪
⎪
∂x
⎫
⎪
⎪
∂
v
⎪
⎪
⎪
=
⎬
⎨ ∂y
⎬ ≈ B d
⎪
⎪
⎪
⎭
⎪ ∂u
∂v ⎪
⎪ ∂y + ∂x ⎪
⎩
⎭
⎡∂N1(x,y)
∂N2(x,y)
0
⎢
∂x
⎢ ∂x
∂N1(x,y)
0
B=⎢ 0
⎢
∂y
⎢∂N (x,y) ∂N (x,y) ∂N (x,y)
1
2
⎢ 1
∂x
∂y
⎣⎢ ∂y
3
a1 = x2 y3 − x3 y 2
a2 = x3 y1 − x1 y3
a3 = x1 y 2 − x2 y1
b1 = y 2 − y3
b2 = y3 − y1
b3 = y1 − y 2
c1 = x3 − x2
c2 = x1 − x3
c3 = x2 − x1
t
k = ∫ e B D B dV
T
V
Since B is constant
A
T
V
0
y1 ⎤
y 2 ⎥⎥
y3 ⎦⎥
Element stiffness matrix
T
∂N3(x,y)
∂x
u3
u2
2 (x2,y2)
k = B D B∫ e dV = B D B At
⎤
⎥
⎡b1 0 b2 0 b3 0 ⎤
⎥
∂N2(x,y)
∂N3(x,y)⎥ 1 ⎢
0
= ⎢ 0 c1 0 c2 0 c3 ⎥⎥
∂y
∂y ⎥ 2A
⎢⎣c1 b1 c2 b2 c3 b3 ⎥⎦
∂N2(x,y) ∂N3(x,y) ∂N3(x,y)⎥
⎥
∂x
∂y
∂x ⎦⎥
0
u
⎡1 x 1
1
A = area of triangle = det ⎢⎢1 x 2
2
⎣⎢1 x 3
v (x,y)≈ N1(x,y)v1 + N2(x,y)v2 + N3(x,y) v3
Approximation of displacements
(x3,y3)
x
u (x,y)≈ N1(x,y)u1 + N2(x,y)u2 + N3(x,y)u3
• 3 nodes per element
• 2 dofs per node (each node can move in x- and y- directions)
• Hence 6 dofs per element
⎧u (x, y)⎫ ⎡ N1
u=⎨
⎬=⎢
⎩ v (x, y)⎭ ⎣ 0
(x,y)
where
u3
v
a1 + b1x + c1 y
2A
a + b x + c2 y
N2 = 2 2
2A
a + b x + c3 y
N3 = 3 3
2A
N1 =
t=thickness of the element
A=surface area of the element
Element nodal load vector
f = ∫ e N X dV + ∫ e N T S dS
V
ST
T
f
T
b
f
S
2
Class exercise
Class exercise
For the CST shown below, compute the vector of nodal loads due to surface traction
f
l1−3e
1
= ∫ e N T S dS
T
S
fS2y
fS2x
2
l1−3e
fS3y
y
fS2y
f S = t∫
fS2x
fS3x
2
3
(0,0)
(1,0)
py=-1
N
fS3x
3
py=-1
T
along 2 −3
T S dS
N
T
along 2 −3
T S dS
⎧0⎫
TS =⎨ ⎬
⎩− 1⎭
fS3y
y
ST
1
f S = t∫
x
The only nonzero nodal loads are
1
f S2 y = t ∫
x1= 0
f S3 y = t ∫
x =0
N 2 along 2 −3 p y dx
N 3 along 2−3 p y dx
a2 + b2 x (x3 y1 − x1 y3 ) + ( y3 − y1 )x
⎡ a + b x + c2 y ⎤
N 2 along 2−3 = ⎢ 2 2
⎥ = 2A =
2A
2A
⎣
⎦ y =0
y1 − y1 x
y1 (1 − x)
y (1 − x)
=
= 1
⎡1 x1 y1 ⎤
⎡1 x1 y1 ⎤ y1 ( x3 − x2 )
det ⎢⎢1 x 2 y2 ⎥⎥ det ⎢⎢1 x 2 0 ⎥⎥
⎣⎢1 x 3 y3 ⎥⎦
⎣⎢1 x 3 0 ⎦⎥
(can you derive this simpler?)
= 1− x
=
x
4-noded rectangular element with edges parallel to the
coordinate axes:
1
⇒ f S2 y = t ∫
x=0
N 2 along 2−3 p y dx
(x4,y4)
4
3 (x3,y3)
1
= t ∫ (1 − x)(−1) dx
4
x =0
=−
v
t
2
u (x,y) ≈ ∑Ni (x,y)ui
u
2b
(x,y)
y
Now compute
1
f S3 y = t ∫
x =0
i =1
1
(x1,y1)
N 3 along 2−3 p y dx
i =1
4
v (x, y) ≈ ∑ N i (x, y)v i
2a
2 (x2,y2)
x
• 4 nodes per element
• 2 dofs per node (each node can move in x- and y- directions)
•8 dofs per element
3
Generation of N1:
At node 1
y
l1 ( x ) =
x − x2
x1 − x 2
has the property
l1(y)
3
4
l1 ( x 2 ) = 0
2b
1
N1
2
2a
l1 ( x1 ) = 1
Similarly
1
y − y4
y1 − y 4
1
(x − x 2 )( y − y 4 )
4 ab
1
(x − x1 )( y − y 3 )
N2 = −
4 ab
1
(x − x 4 )( y − y 2 )
N3 =
4 ab
1
(x − x 3 )( y − y1 )
N4 = −
4 ab
N1 =
has the property
l1 ( y 1 ) = 1
x
l1 ( y 4 ) = 0
l1(x)
1
l1 ( y ) =
Using similar arguments, choose
Hence choose the shape function at node 1 as
⎛ x − x2
N 1 = l1 ( x )l1 ( y ) = ⎜⎜
⎝ x1 − x 2
⎞⎛ y − y 4
⎟⎟⎜⎜
⎠⎝ y1 − y 4
⎞
1
⎟⎟ =
(x − x 2 )( y − y 4 )
⎠ 4 ab
Properties of the shape functions:
1. The shape functions N1, N2 , N3 and N4 are bilinear functions
of x and y
2. Kronecker delta property
⎧ 1 at node ' i '
N i ( x, y ) = ⎨
⎩0 at other nodes
3. Completeness
4
∑
i =1
4
∑
3. Along lines parallel to the x- or y-axes, the shape functions
are linear. But along any other line they are nonlinear.
4. An element shape function related to a specific nodal point is
zero along element boundaries not containing the nodal point.
5. The displacement field is continuous across elements
6. The strains and stresses are not constant within an element
nor are they continuous across element boundaries.
Ni =1
N i xi = x
i =1
4
∑
i =1
N i yi = y
4
The strain-displacement relationship
Computation of the terms in the stiffness matrix of 2D elements (recap)
⎧εx ⎫
⎪ ⎪
ε = ⎨ε y ⎬
⎪γ xy ⎪
⎩ ⎭
v4
⎧u1 ⎫
⎪v ⎪
⎤⎪ 1 ⎪
⎡ ∂N1(x, y)
∂N3(x, y)
∂N 2 (x, y)
∂N 4 (x, y)
0
0
0
0
⎥⎪u 2 ⎪
⎢
∂x
∂x
∂x
⎥⎪ ⎪
⎢ ∂x
∂N3(x, y)
∂N1(x, y)
∂N 2 (x, y)
∂N 4 (x, y)⎥⎪v 2 ⎪
0
0
0
=⎢ 0
⎨ ⎬
⎢
∂y
∂y
∂y
∂y ⎥⎪u 3 ⎪
⎢ ∂N (x, y) ∂N (x, y) ∂N (x, y) ∂N (x, y) ∂N (x, y) ∂N (x, y) ∂N (x, y) ∂N (x, y)⎥
3
3
1
2
2
4
4
⎥⎪v3 ⎪
⎢ 1
y
∂x
∂y
∂x
∂y
∂x
∂y
∂x ⎥⎪ ⎪
⎣⎢ ∂
⎦⎪u 4 ⎪
B
⎪v ⎪
⎩ 4⎭
y3 − y
y − y2
y1 − y 0 ⎤
0
0
0
⎡y − y4
1 ⎢
B=
x − x2
x1 − x
x − x4
x3 − x⎥⎥
0
0
0
0
⎢
4ab
⎣⎢ x − x2 y − y4 x1 − x y3 − y x − x4 y − y2 x3 − x y1 − y⎦⎥
Notice that the strains (and hence the stresses) are NOT constant within an element
v3
4
u4
The B-matrix (strain-displacement) corresponding to this element is
3
u3
u1
v
y
v1 (x,y)
v2
u
2
1 u1
u2
⎡ ∂N (x,y)
⎢ 1
∂x
⎢
⎢
0
⎢
⎢
⎢ ∂N1 (x,y)
⎢
∂y
⎣⎢
v1
u2
v2
0
∂N 2 (x,y)
∂x
0
∂N1 (x,y)
∂y
∂N1 (x,y)
∂x
0
∂N 2 (x,y)
∂y
∂N2 (x,y)
∂y
∂N2 (x,y)
∂x
u3
∂N 3 (x,y)
∂x
0
∂N 3 (x,y)
∂y
v3
0
∂N 3 (x,y)
∂y
∂N 3 (x,y)
∂x
u4
v4
∂N4 (x,y)
∂x
⎤
0
⎥
⎥
∂N4 (x,y) ⎥
⎥
∂y
⎥
∂N4 (x,y) ⎥
⎥
∂x
⎦⎥
0
∂N4 (x,y)
∂y
x
We will denote the columns of the B-matrix as
B u1
⎡
⎤
⎡ ∂ N 1 (x,y) ⎤
⎢
⎥
0
⎢
⎥
⎢
⎥
x
∂
⎢
⎥
⎢ ∂ N 1 (x,y) ⎥ ; and so on...
0
=⎢
⎥ ; B v1 = ⎢
⎥
y
∂
⎢ ∂ N (x,y) ⎥
⎢
⎥
1
⎢
⎥
⎢ ∂ N 1 (x,y) ⎥
∂y
⎣⎢
⎦⎥
⎢⎣
⎥
∂x
⎦
The stiffness matrix corresponding to this element is
k = ∫ e B D B dV
T
which has the following form
V
u1
v1
⎡ k11
⎢k
⎢ 21
⎢ k 31
⎢
k
k = ⎢ 41
⎢ k 51
⎢
⎢ k 61
⎢k
⎢ 71
⎢⎣ k 8 1
u2
v2
u3
v3
u4
v4
k18 ⎤
k 2 8 ⎥⎥
k 38 ⎥
⎥
k 48 ⎥
k 58 ⎥
⎥
k 68 ⎥
k 78 ⎥
⎥
k 8 8 ⎥⎦
k12
k 22
k13
k 23
k14
k 24
k15
k 25
k16
k 26
k17
k 27
k 32
k 42
k 33
k 43
k 34
k 44
k 35
k 45
k 36
k 46
k 37
k 47
k 52
k 62
k 53
k 63
k 54
k 64
k 55
k 65
k 56
k 66
k 57
k 67
k 72
k 82
k 73
k 83
k 74
k 84
k 75
k 85
k 76
k 86
k 77
k 87
u1
v1
u2
Notice that these formulae are quite general (apply to all kinds
of finite elements, CST, quadrilateral, etc) since we have not
used any specific shape functions for their derivation.
v2
u3
v3
u4
v4
The individual entries of the stiffness matrix may be computed as follows
k11 = ∫ e Bu1 D Bu1 dV; k12 = ∫ e Bu1 D Bv1 dV; k13 = ∫ e Bu1 D Bu2 dV,...
T
T
V
V
T
V
k 21 = ∫ e Bv1 D Bu1 dV; k21 = ∫ e Bv1 D Bv1 dV;.....
T
V
T
V
5
Example
1000 lb
Realize that this is a plane stress problem and therefore we need to use
300 psi
y
3
4
Thickness (t) = 0.5 in
E= 30×106 psi
ν=0.25
2 in
1
2
D=
⎤
⎡
1 ν
0 ⎥ ⎡3.2 0.8 0 ⎤
E ⎢
⎢ν 1
0 ⎥ = ⎢⎢0.8 3.2 0 ⎥⎥ ×10 7 psi
2
1 −ν ⎢
1 −ν ⎥
0 1.2⎥⎦
⎥ ⎢⎣ 0
⎢0 0
2 ⎦
⎣
Write down the shape functions
x
3 in
1
(x − x2 )( y − y 4 ) = ( x − 3)( y − 2)
4 ab
6
1
N2 = −
(x − x1 )( y − y3 ) = − x ( y − 2)
4 ab
6
1
(x − x4 )( y − y 2 ) = xy
N3 =
4 ab
6
1
(x − x3 )( y − y1 ) = − ( x − 3) y
N4 = −
4 ab
6
N1 =
(a) Compute the unknown nodal displacements.
(b) Compute the stresses in the two elements.
This is exactly the same problem that we solved in last class, except
now we have to use a single 4-noded element
x
y
0
0
3
0
3
2
0
2
Compute only the relevant columns of the B matrix
We have 4 nodes with 2 dofs per node=8dofs. However, 5 of these are fixed.
The nonzero displacements are
u2
u3
v3
Hence we need to solve
u2
u3
v3
⎡ k11
⎢k
⎢ 21
⎣⎢ k 31
k12
k 22
k 32
k13 ⎤ ⎧u 2 ⎫ ⎧ 0 ⎫
⎪ ⎪ ⎪ ⎪
k 23 ⎥⎥ ⎨u 3 ⎬ = ⎨ 0 ⎬
k 33 ⎦⎥ ⎪⎩ v3 ⎪⎭ ⎪⎩ f 3 y ⎪⎭
⎧ ∂N 2 ⎫ ⎧ ( 2 − y ) ⎫
⎪ ∂x ⎪ ⎪ 6 ⎪
⎪
⎪
⎪ ⎪
B u2 = ⎨ 0 ⎬ = ⎨ 0 ⎬
⎪ ∂N 2 ⎪ ⎪ − x ⎪
⎪ ∂y ⎪ ⎪
6 ⎪⎭
⎩
⎭ ⎩
B u3
⎧ ∂N 3 ⎫ ⎧ y ⎫
⎪ ∂x ⎪ ⎪ 6 ⎪
⎪ ⎪ ⎪
⎪
= ⎨ 0 ⎬ = ⎨0⎬
⎪ ∂N 3 ⎪ ⎪ x ⎪
⎪ ∂y ⎪ ⎪ 6 ⎪
⎭ ⎩ ⎭
⎩
B v3
⎧
⎫ ⎧ ⎫
⎪ 0 ⎪ ⎪0⎪
⎪
⎪
⎪ ∂N 3 ⎪ ⎪⎪ x ⎪⎪
=⎨
⎬=⎨ ⎬
⎪ ∂y ⎪ ⎪ 6 ⎪
∂
N
⎪ 3⎪ ⎪y⎪
⎪⎩ ∂x ⎪⎭ ⎩⎪ 6 ⎭⎪
Need to compute only the relevant terms in the stiffness matrix
k11 = ∫ e Bu2 D Bu2 dV; k12 = ∫ e Bu2 D Bu3 dV; k13 = ∫ e Bu2 D Bv3 dV
T
V
T
V
T
V
k 21 = ∫ e Bu3 D Bu2 dV; k22 = ∫ e Bu3 D Bu3 dV; k13 = ∫ e Bu3 D Bv3 dV
T
V
T
V
T
V
k31 = ∫ e Bv3 D Bu2 dV; k 22 = ∫ e Bv3 D Bu3 dV; k13 = ∫ e Bv3 D Bv3 dV
T
V
T
V
T
V
6
How do we compute f3y
k11 = ∫ e Bu2 D Bu2 dV
T
V
3
2
= 0.5 ∫
∫
x=0 y=0
2− y
⎡
8
7
5 2⎤
⎢(0.1067 × 10 − 0.533 × 10 )( 6 ) + 3.33 × 10 x ⎥ dxdy
⎣
⎦
= 0.656 × 107
f 3 y = − 1000 + f S 3 y
f S3 y = t ∫
3
x =0
Similarly compute the other terms
N 3 edge = N 3
4 −3
N3
along
edge 3 − 4
= ( 0 .5)( − 300 ) ∫
( − 300 ) dx
3
x =0
x
dx
3
4
y =2
x
⎡ xy ⎤
=⎢ ⎥ =
⎣ 6 ⎦ y =2 3
3
3
= − 150 ×
2
= − 225 lb
⇒ f 3 y = − 1000 + f S 3 y = − 1225 lb
How about a 9-noded rectangle?
Corner nodes
y 5
a
a
2
b
6
b
3
9
7
⎡ x(a + x) ⎤ ⎡ y(b + y) ⎤
⎡ x(a − x) ⎤ ⎡ y(b + y) ⎤
N1 = ⎢
2
2
⎥⎢
⎥ N 2 = ⎢− 2a 2 ⎥ ⎢ 2b 2 ⎥
⎣ 2a ⎦ ⎣ 2b ⎦
⎣
⎦⎣
⎦
⎡ x(a − x) ⎤ ⎡ y(b − y ) ⎤
⎡ x(a + x) ⎤ ⎡ y(b − y) ⎤
−
=
−
N 3 = ⎢−
N
4
⎢⎣ 2a 2 ⎥⎦ ⎢⎣
2a 2 ⎦⎥ ⎣⎢
2b 2 ⎦⎥
2b 2 ⎥⎦
⎣
1
8
4
x
Midside nodes
2
2
⎡ a 2 − x 2 ⎤ ⎡ y(b + y) ⎤
⎡ x (a − x ) ⎤ ⎡ b − y ⎤
N5 = ⎢
⎥⎢
2
2
⎥ N 6 = ⎢⎣− 2a 2 ⎥⎦ ⎢ b 2 ⎥
⎣ a
⎦ ⎣ 2b ⎦
⎣
⎦
2
2
⎡ a 2 − x 2 ⎤ ⎡ y(b − y) ⎤
⎡ x( a + x) ⎤ ⎡ b − y ⎤
−
N7 = ⎢
=
N
8
⎥⎢
2
⎢⎣ 2a 2 ⎥⎦ ⎢ b 2 ⎥
2b 2 ⎦⎥
⎣ a
⎦⎣
⎣
⎦
Center node
⎡ a 2 − x2 ⎤ ⎡ b2 − y 2 ⎤
N9 = ⎢
⎥⎢
⎥
2
2
⎣⎢ a
⎦⎥ ⎣⎢ b
⎦⎥
Question: Can you generate the shape functions of a 16-noded rectangle?
Note: These elements, whose shape functions are generated by multiplying the
shape functions of 1D elements, are said to belong to the “Lagrange” family
7